Problem: The increasing sequence $3, 15, 24, 48, \ldots\,$ consists of those positive multiples of $3 $ that are one less than a perfect square. What is the remainder when the $1994$th term of the sequence is divided by $1000$? Solution: So let me tell you an embarrassing story about this problem. It took me almost an hour to figure out where my computation went wrong despite my solid approach within the first five minutes. I had the audacity to email Desmos and tell them that their calculator wasn't working because I was so convinced something was wrong. Well, something WAS wrong, and I wrote $977$ instead of $997$. I belong in math jail for that one. Anyway... the question narrows down to realizing that the sequence is generated by $n^2-1$ where $n \equiv 1,2 \pmod 3$. If $n \equiv 0 \pmod 3$ then $n^2-1$ is not divisible by $3$. So we want to find the $\frac{1994}{2} = 997$th positive integer satisfying $n \equiv 1 \pmod 3$. Or do we? Notice that the sequence starts with $2^2-1=3$ and this is the second such positive integer that is one modulo three. So we want to find the $998$th positive integers that is one modulo three. This is given by $(1+997\cdot 3)^2 - 1$. We want to find this number $\pmod {1000}$ so we see that $(1+997\cdot 3)^2-1 \equiv (-8)^2 - 1 \pmod {1000} = 63.$